Geeta's A.P. is -5, -2, 1,4, ...
Here, first term (a₁) = -5 and common difference (d₁) = -2 + 5 = 3
Similarly, Madhuri's A.P. is 187, 184, 181, ...
Here first term (a₂) = 187 and common difference (d₂) = 184 - 187 = -3
(i) (b): t₃₄ = a₂ + 33d₂ = 187 + 33(-3) = 88
(ii) (d): Required sum = 3 + (-3) = 0
(iii) (a): t₁₉ = a₁ + 18d₁ = (-5) + 18(3) = 49
(iv) (a) : \(S_{10}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]=\frac{10}{2}[2(-5)+9(3)]=85\)
(v) (b): Let n^th terms of the two A.P:s be equal.
\(\therefore\) -5 + (n - 1)3 = 187 + (n - 1)(-3)
\(\Rightarrow\) 6(n - 1) = 192 \(\Rightarrow\) n = 33

(i) (d)
(ii) (b)
(iii) (a): x(x + 3) + 7 = 5x - 11
\(\Rightarrow x^{2}+3 x+7=5 x-11\)
\(\Rightarrow x^{2}-2 x+18=0 \) is a quadratic equation.
\((b) (x-1)^{2}-9=(x-4)(x+3)\)
\(\Rightarrow x^{2}-2 x-8=x^{2}-x-12\)
\(\Rightarrow x-4=0\) is not a quadratic equation.
\((c) x^{2}(2 x+1)-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}+x^{2}-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}-4 x^{2}+6=0\) is not a quadratic equation.
\((d) x(x-1)(x+7)=x(6 x-9)\)
\(\Rightarrow x^{3}+6 x^{2}-7 x=6 x^{2}-9 x\)
\(\Rightarrow x^{3}+2 x=0\)   is not a quadratic equation.
(iv) (d)
(v) (d)

(i) (b): Roots of the quadratic equation are 2 and -3.
\(\therefore\) The required quadratic equation is 
\((x-2)(x+3)^{n}=0 \Rightarrow x^{2}+x-6=0\)
(ii) (a): We have, 2x² + kx + 1 = 0
Since, -1/2 is the root of the equation, so it will satisfy the given equation
\(\therefore \quad 2\left(-\frac{1}{2}\right)^{2}+k\left(-\frac{1}{2}\right)+1=0 \Rightarrow 1-k+2=0 \Rightarrow k=3\)
(iii) (d): If the roots of the quadratic equations are opposites to each other, then coefficient of x (sum of roots) is 0.
So, both (a) and (b) have the coefficient of x = 0.
(iv) (c): The given equation is (x - 2)² + 19 = 0
\(\Rightarrow x^{2}-4 x+4+19=0 \Rightarrow x^{2}-4 x+23=0\)
(v) (b): If one root of a quadratic equation is irrational, then its other root is also irrational and also its conjugate i.e., if one root is p +.\(\sqrt(q)\) then its other root is p -.\(\sqrt(q)\).

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600.
(ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500.
(iii) (c) : At x-axis, y = 0
\(\therefore\)  At y = 0, 2x + y = 1600 becomes 2x = 1600
\(\Rightarrow\) x = 800
\(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0).
(iv) (d) : At y-axis, x = 0
\(\therefore\) 2x + Y = 1500 \(\Rightarrow\) y = 1500
\(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500).
(v) (b): We have, 2x + y = 1600 and 2x + y = 1500
Since \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\)
\(\therefore\) System of equations have no solution.
\(\therefore\) Lines are parallel.

(i) (a): At x-axis, y = 0
\(\therefore\) 2x + 4y = 8 \(\Rightarrow\) x = 4
At y-axis, x = 0
\(\therefore\) 2x + 4y = 8 \(\Rightarrow\) Y = 2
\(\therefore\) Required coordinates are (4, 0), (0, 2).
(ii) (c): At x-axis, y = 0
\(\therefore\) 3x + 6y = 18 \(\Rightarrow\) 3x = 18 \(\Rightarrow\) x = 6
At y-axis, x = 0
\(\therefore\) 3x + 6y = 18 \(\Rightarrow\) 6y = 18 \(\Rightarrow\) Y = 3
\(\therefore\) Required coordinates are (6, 0), (0, 3).
(iii) (d): Since, lines are parallel. So, point of intersection of these lines does not exist.
(iv) (a)
(v) (a): Since the lines are parallel.
\(\therefore\) These equations have no solution i.e., the given system of linear equations is inconsistent.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850
(ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900
(iii) (c) : We have 2x + 3y = 850 .........(i)
and 3x + 2y = 900 .........(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get
5y = 750 \(\Rightarrow\) Y = 150
Thus, price of one Physics book is Rs 150.
(iv) (d): From equation (i) we have, 2x + 3 x 150 = 850
\(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200
Hence, cost of one Mathematics book = Rs 200
(v) (a): From above, we have
\(a_{1} =2, b_{1}=3, c_{1}=-850 \)
\(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\)
\(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus system of linear equations has unique solution.

(i) (a) : Suppose two cars meet at point Q. Then,
Distance travelled by car X = A Q,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
\(\therefore\) Distance travelled by car X in 9 hours = 9x km
\(\Rightarrow\) AQ=9x
Distance travelled by car Y in 9 hours = 9y km
\(\Rightarrow\) BQ=9y

Clearly, AQ - BQ = AB
\(\Rightarrow\) 9x - 9y = 90
\(\Rightarrow\) x-y =10
(ii) (c): Suppose two cars meet at point P. Then
Distance travelled by car X = AP and
Distance travelled by car Y = BP.
In this case, two cars meet in 917 hours.
\(\therefore\) Distance travelled by car X in 9/7 hours = \(\frac {9}{7}\) x km
\(\Rightarrow A P=\frac{9}{7} x\)
Distance travelled by car Y in 9/7 hours \(\frac {9}{7}\) y km
\(\Rightarrow B P=\frac{9}{7} y\)
Clearly, AP + BP = AB
\(\Rightarrow \quad \frac{9}{7} x+\frac{9}{7} y=90 \Rightarrow \frac{9}{7}(x+y)=90 \Rightarrow x+y=70\)
(iii) (b): We have x - y = 10
\(\Rightarrow x+y=70\)
Adding equations (i) and (ii), we get
2x = 80 \(\Rightarrow\) x = 40
Hence, speed of car X is 40 km/hr.
(iv) (c): We have x - y = 10
\(\Rightarrow\) 40 - Y = 10 \(\Rightarrow\) Y = 30
Hence, speed of car y is 30 km/hr.
(v) (d)

(i) (d): For finding zeroes, check whether \(x^{3}-4 x^{2}-7 x+10\) is 0 for given zeroes
Let p(x) = x³ - 4x² - 7x + 10. Then, Clearly p( -2) = p(1) = p(5) = 0 So, the zeroes are -2, 1 and 5.
(ii) (d): Here \(\alpha=\frac{-1}{2}, \beta=-2 \text { and } \gamma=5\)
\(\therefore\) Sum of product of zeroes taken two at a time 
\(=\alpha \beta+\beta \gamma+\gamma \alpha \)
\(=\left(\frac{-1}{2}\right)(-2)+(-2)(5)+(5)\left(\frac{-1}{2}\right)=1-10-\frac{5}{2}=\frac{-23}{2}\)
(iii) (d): Consider \(x^{3}-x^{2}+5 x-1\)
Sum of zeroes = 1 = Product of zeroes
Now, consider x³ - 4x
Sum of zeroes = 0 = Product of zeroes.
(iv) (b): Let a, a, a, be the zeroes of the cubic polynomial. [\(\because\) All zeroes are same]
Then, a3 = 1 => a = 1 [Using given options]
So, the required polynomial is \((x-1)^{3}=x^{3}-3 x^{2}+3 x-1\)
(v) (a): Clearly x = 1 and x = 2 are the zeroes of given polynomial, both of which satisfies \(x^{3}-5 x^{2}+8 x-4\)
 

(i) (b)
(ii) (d): Since, the parabola intersects the x-axis at -3 and 4. So, zeroes of the polynomial are -3 and 4.
(iii) (c): Let \(f(x)=x^{2}-x-12\)
\(=x^{2}-4 x+3 x-12=(x+3)(x-4) \)
\(\text { Consider } f(x)=0 \Rightarrow(x+3)(x-4)=0 \Rightarrow x=4,-3\)
(iv) (b): Sum of zeroes \(=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}} \)
\(=-\frac{(-9)}{4}=\frac{9}{4}\)
(v) (c): Product of zeroes \(=\frac{20}{1}=20\)
\(\therefore\) Reciprocal of product of zeroes \(=\frac{20}{1}\)

(i) (b):Put \(10 x^{2}-x-3=0\)
\(\Rightarrow 10 x^{2}-6 x+5 x-3=0 \Rightarrow(2 x+1)(5 x-3)=0 \)
\(\Rightarrow \quad x=\frac{-1}{2} \text { or } \frac{3}{5}\)
\(\text { Thus, the zeroes are } \frac{3}{5} \text { and } \frac{-1}{2} \text { . }\)
(ii) (a): The zeroes of the polynomial are the points where its graph intersect the x-axis.
(iii) (d)
(iv) (d)
(v) (c): Product of zeroes \(=\frac{8 \sqrt{3}}{\sqrt{3}}=8\).

(i) (b): The shape of the path of the soccer ball is a parabola.
(ii) (c): The axis of symmetry of the given curve is a line parallel to y-axis.
(iii) (a): The zeroes of the polynomial, represented in the given graph, are -2 and 7, since the curve cuts the x-axis at these points.
(iv) (d):A polynomial having zeroes -2 and -3 is \(p(x)=x^{2}-(-2-3) x+(-2)(-3)=x^{2}+5 x+6\)
(v) (c): We have \(f(x)=(x-3)^{2}+9\)
\(\text { Now, } 9=(x-3)^{2}+9 \)
\(\Rightarrow(x-3)^{2}=0 \Rightarrow x-3=0 \Rightarrow x=3\)

(i) (b): Since the graph of the polynomial intersect the x-axis at \(x=\frac{1}{2}, \frac{-7}{2}\), therefore required zeroes of the polynomial are \(\frac{1}{2} \text { and } \frac{-7}{2}\)
(ii) (d): \(\because \frac{1}{2} \text { and } \frac{-7}{2}\) are the zeroes of the polynomial.
So, at \(x=\frac{1}{2}, \frac{-7}{2}\) the value of the polynomial will be 0.
From options, required polynomial is
p(x) = -4x² - 12x + 7
(iii) (b) : we have, \(p(x)=-4 x^{2}-12 x+7\)
\(\therefore \quad p(3)=-4(3)^{2}-12(3)+7=-36-36+7=-65 \)
(iv) (c): Here \(f(x)=x^{2}+2 x-8 \text { and } \alpha, \beta \text { are its zeroes. }\)
\(\therefore \quad \alpha+\beta=-2 \text { and } \alpha \beta=-8 \)
\(\text { Now, } \alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2} \)
\(=\left((\alpha+\beta)^{2}-2 \alpha \beta\right)^{2}-2(\alpha \beta)^{2} \)
\(=\left[(-2)^{2}-2(-8)\right]^{2}-2(-8)^{2} \)
\(=[4+16]^{2}-2(-8)^{2}=(20)^{2}-2(64) \)
\(=400-128=272\)
(v) (a): We have sum of zeroes = 0 and product of zeroes = \(\sqrt(7)\)
So, required polynomial .\(=k\left(x^{2}-0 \cdot x+\sqrt{7}\right) \)
\(=k\left(x^{2}+\sqrt{7}\right)\)

(i) (b): Since, the given graph is parabolic is shape, therefore it will represent a quadratic polynomial.
[\(\therefore\) Graph of quadratic polynomial is parabolic in shape 1
(ii) (c): Since, the graph cuts the x-axis at -1, 5. So the polynomial has 2 zeroes i.e., -1 and 5.
(iii) (a) : Sum of zeroes = -1 + 5 = 4
(iv) (c): Since a and β are zeroes of the given polynomial and β > a
\(\therefore\)a = - 1 and β = 5.
\(\therefore|8 \alpha+\beta|=|8(-1)+5|=|-8+5|=|-3|=3 .\)
(v) (d): Since the zeroes of the given polynomial are - 1 and 5.
\(\therefore\) Required polynomial p(x)
= k{ x² -(-1 + 5)x + (-1)(5)} = k(.x² - 4x - 5)
For k = -1, we get
p(x) = -.x² + 4x + 5, which is the required polynomial.

(i) (c): Here, the simplest form of given options are
125/441 = 5³/(3² x 7²), 77/210 = 11/(2 x 3 x 5),
15/1600 = 3/(2⁶ x 5) Out of all the given options, the denominator of option (c) alone has only 2 and 5 as factors. So, it is a terminating decimal.
(ii) (b): 23/(2³ x 5²) = 23/200 = 0.115
(iii) (a): 441/(2² x 5⁷ x 7²) = 9/(2² x 5⁷), which is a terminating decimal.
(iv) (d): The fraction form of a non-terminating recurring decimal will have at least one prime number other than 2 and 5 as its factors in denominator. So, p can take either of 3, 7 or 15.
(v) (a): Here denominator has only two prime factors i.e., 2 and 5 and hence it is a terminating decimal.

(i) (b): Here 80 = 2⁴ x 5, 85 = 17 x 5
and 90 = 2 x 3² x 5
L.C.M of 80, 85 and 90 = 2⁴ x 3 x 3 x 5 x 17 = 12240
Hence, the minimum distance each should walk when they at first time is 12240 cm.
(ii) (c): Here 594 = 2 x 3³ x 11 and 189 = 3³ x 7
HCF of 594 and 189 = 3³= 27
Hence, the maximum number of columns in which they can march is 27.
(iii) (c) : Here 768 = 2⁸ x 3 and 420 = 2² x 3 x 5 x 7
HCF of 768 and 420 = 2² x 3 = 12
So, the container which can measure fuel of either tanker exactly must be of 12litres.
(iv) (b): Here, Length = 825 ern, Breadth = 675 cm and Height = 450 cm
Also, 825 = 5 x 5 x 3 x 11 , 675 = 5 x 5 x 3 x 3 x 3 and 450 = 2 x 3 x 3 x 5 x 5
HCF = 5 x 5 x 3 = 75
Therefore, the length of the longest rod which can measure the three dimensions of the room exactly is 75cm.
(v) (a): LCM of 8 and 12 is 24.
\(\therefore \)The least number of pack of pens = 24/8 = 3
\(\therefore \)The least number of pack of note pads = 24/12 = 2